Just list the simple events, there aren't that many of them.
1) Four friends, 2 females and 2 males are playing contract bridge. Partners are randomly assigned. What is the probability that two females will be partners for the first game?
I did:
P(2 females) = P(no 2 females)
= P(6C2)
(8C2)
= 15
28
and i got it wrong....dont know what i did wrong.
Hello, questiongirl!
This is easier/harder than we'd expect . . .
Four friends, 2 females and 2 males are playing contract bridge.
Partners are randomly assigned.
What is the probability that two females will be partners for the first game?
Suppose the four friends are: .
There are only three assignments for partners:
. .
Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Choose a female.
. . Either one, it makes no difference.
Now what is the probability that her partner is the other female?
There are 3 choices: .
Therefore: .